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Table 20-2
Minimum Operating Frequencies [MHz]
Number of allocated message
2)
objects MO
, without TTCAN
functionality
16 MO, without TTCAN
32 MO, without TTCAN
64 MO, without TTCAN
128 MO, without TTCAN
1) In the case of 15 time quanta, the minimum operating frequency required is 15MHz.
2) Only those message objects have to be taken into account that are allocated to a CAN node. The unallocated
message objects have no influence on the minimum operating frequency.
The baud rate generation of the MultiCAN being based on
chosen carefully to allow correct CAN bit timing. The required value of
integer multiple (n) of the CAN baud rate multiplied by the number of time quanta per
CAN bit time. For example, to reach 1 Mbit/s with 20 tq per bit time, possible values of
are given by formula [n × 20] MHz, with n being an integer value, starting at 1.
In order to minimize jitter, it is not recommended to use the fractional divider mode for
high baud rates.
Additionally, for correct operation of the MultiCAN, the following conditions have to be
fulfilled,
[
(
Baudrate
=
8 T
max
also
<
NBTR.SJW NBTR.TSEG1
As an example, when
Baudrate
=[(8x50ns) + (8x50ns) + (4 x 1 x 50ns)] = 1000ns = 1MBaud
max
Table 20-3
below illustrates the minimum CAN module timer clock
f
Control Clock
CLC
higher baudrate is desired, the values need to be calculated as per
User's Manual
MultiCAN, V2.24
1 CAN
node
active
12
15
21
40
×
)
(
×
+
8 T
CAN
CLC
f
f
=
=20MHz, No of active CAN nodes =1,
CAN
CLC
thats required to support a Baurate Generation of 500KBaud. If a
Controller Area Network Controller (MultiCAN)
2 CAN
3 CAN
nodes
nodes
active
active
19
26
23
30
28
37
45
50
)
(
×
+
4 No. of active CAN nodes T
20-18
1)
4 CAN
nodes
active
33
37
46
55
f
, this frequency has to be
A
f
is given by an
A
×
CLC
f
CAN
Equation
V1.0, 2011-12
TC1728
5 CAN
nodes
active
40
44
53
61
f
A
)
]
(20.1)
and Module
(20.1).
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